using a = bcosC+ccosB,By Cauchy,(b^2+c^2)(cos^2C+cos^2B)≥a^2.

$(sin^2\alpha+sin^2\beta)(cos^2\alpha+cos^2\beta)\ge(sin\alpha cos\beta+sin\beta cos\alpha)=sin\gamma$ $a=2Rsin\alpha$...

qweDota wrote: Let $\alpha,\beta,\gamma$ measures of angles of opposite to the sides of triangle with measures $a,b,c$ respectively.Prove that $$2(cos^2\alpha+cos^2\beta+cos^2\gamma)\geq \frac{a^2}{b^2+c^2}+\frac{b^2}{a^2+c^2}+\frac{c^2}{a^2+b^2}$$ https://artofproblemsolving.com/community/c6h75424p2731051

$cos \alpha =\frac {p}{b}$ $cos \beta =\frac {k}{c}$ Where $a, b, c$ are sides of a triangle And $a=k+p$,$k, p\in\mathbb{R^+}$ $cos^2\alpha +cos^2\beta =\frac{p^2}{b^2}+\frac{k^2}{c^2}\ge \frac{(k+p)^2}{b^2+c^2}$ $=\frac{a^2}{b^2+c^2}$ Similarly for other sides and then add them up which does the work.

This inequality was already proposed in 2011. http://artofproblemsolving.com/community/c6h423450p2394999

mudok wrote: This inequality was already proposed in 2011. http://artofproblemsolving.com/community/c6h423450p2394999 This inequality was already proposed in 2008. Maybe earlier...

The Cosine law and Cauchy-Schwarz in Engel form yield $\cos^2\alpha + \cos^2\beta = \frac{(b^2+c^2-a^2)^2}{4b^2c^2} + \frac{(a^2+c^2-b^2)^2}{4a^2c^2} \geq \frac{(2c^2)^2}{4c^2(a^2+b^2)} = \frac{c^2}{a^2+b^2}$. Summing cyclically yields the result.